To solve the problem, we need to find the electromotive force (emf) of the ideal battery in the circuit. Here’s a step-by-step process:

Given:

• Current through $R_6$, $i_6 = 1.45 \, \text{A}$
• Resistors:
  • $R_1 = R_2 = R_3 = 1.80 \, \Omega$
  • $R_4 = 14.6 \, \Omega$
  • $R_5 = 8.44 \, \Omega$
  • $R_6 = 3.79 \, \Omega$

Step 1: Find the total resistance of the circuit.

All the resistors are in series, so the total resistance $R_{\text{total}}$ is the sum of all the resistances: $R_{\text{total}} = R_1 + R_2 + R_3 + R_4 + R_5 + R_6$ Substitute the given values: $R_{\text{total}} = 1.80 + 1.80 + 1.80 + 14.6 + 8.44 + 3.79$ $R_{\text{total}} = 32.23 \, \Omega$

Step 2: Use Ohm's Law to calculate the emf of the battery.

Ohm's Law states that: $V = I \times R$ Where:

• $V$ is the emf of the battery
• $I$ is the current through the circuit
• $R$ is the total resistance

Since the current through $R_6$ is given as $1.45 \, \text{A}$ and all components are in series, the current is the same throughout the circuit. Therefore, the emf is: $\text{emf} = I \times R_{\text{total}} = 1.45 \times 32.23$ $\text{emf} = 46.24 \, \text{V}$

Answer:

The emf of the ideal battery is approximately $46.24 \, \text{V}$.

Would you like any more details on this solution or have any further questions?

Here are some related questions you might find useful:

1. What would happen if one of the resistors was removed from the circuit?
2. How does changing the value of $R_6$ affect the emf?
3. What is the power consumed by each resistor in this circuit?
4. How would the circuit behavior change if the battery was not ideal (had internal resistance)?
5. How is the total current in the circuit affected if an additional resistor is added in parallel with $R_6$?

Tip:

When solving series circuits, remember that the current remains constant throughout all components, while the voltage across each component depends on its resistance.